Function such that $f(x) f(\pi/2 - x) = 1$

I'm looking for functions that are smooth ($C^\infty$) between $0 < x < \pi/2$ that satisfy the equation $$f(x)\, f(\pi/2-x) = 1$$ on the inteverval $0<x<\pi/2$. I know that the constant function $f=1$ satisfies this equation, as well as $f=\tan(x)$. Are these the only solutions?

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It seems that there are infinitely many $C^\infty$ functions that work, so long as the power series at $x=\pi/4$ is consistent with the restrictions coming from taking derivatives of the above expression at $\pi/4$. Each of these power series should correspond to an analytic function that satisfies the above equation in a neighborhood of $x=\pi/4$. So the question seems to be how many analytic solutions are there to the above problem?

• Also $f(x) = -1$ and $-\tan(x)$. Others ...? – Simon S Nov 30 '14 at 20:24
• $\pm\tan x$ fails for $x=0$ – Alice Ryhl Nov 30 '14 at 20:46
• There are infinitely many such functions even assuming continuity and differentiability - just choose $f(x)$ bounded away from zero on $0\leq x\leq\frac\pi4$, with $f(\frac\pi4)=\pm 1$ and an appropriate value for $f'(\frac\pi4)$, and then define $f(y)$ on $\frac\pi4\leq y\leq\frac\pi2$ using $f(y)=\frac1{f(\frac\pi2-y)}$. – Steven Stadnicki Nov 30 '14 at 21:02
• @StevenStadnicki You only define $f(x)$ on $0\le x\le \pi/2$ – Alice Ryhl Nov 30 '14 at 21:10
• @KristofferRyhl oh whoops, I should probably say that it doesn't have to hold at $x=0$ since I want to allow $f$ to diverge as $x\rightarrow \pi/2$. – asperanz Nov 30 '14 at 21:30

Taking the logarithm gives $$\log f(x)+\log f(\pi/2 - x)=0,$$ or $$\log f\left(\pi/4 + (x-\pi/4)\right)=-\log f\left(\pi/4 - (x - \pi/4)\right).$$ That is, $\log f$ needs to be odd under reflection across $\pi/4$. So let $g(x)$ be any odd function (defined at least for $|x|<\pi/4$); then $$f(x)=\exp{g(x - \pi/4)}$$ meets the conditions. If $g(x)$ is chosen to be continuous, smooth, or analytic, then $f(x)$ will be an equally "nice" function. For instance, $f(x)=e^{x-\pi/4}$ is a simple analytic example, or $f(x)=e^{(x-\pi/4)^3}$.
Rescaling the interval to $(0,2),$ we want $f(x)f(2-x)=1$. Assuming this $f$ is smooth, and say $f(1)=1$ [one of the two choices there], let $g(x)$ be its restriction to $[0,1]$, and then on $[1,2]$ we have $f(x)=h(x) \equiv 1/g(2-x).$ The derivative matching is automatic: $$h'(x)=\frac{-1}{g(2-x)^2}g'(2-x)\cdot (-1),$$ which since $g(1)=1$ and $g'(2-1)=g'(1),$ gives a match between the left side derivative of $g$ at $x=1$ and the right side derivative of $h$ at $1.$
So it seems $f$ may be chosen arbitrarily nonzero on $(0,1]$ with $f$ smooth on a on that interval (derivative from the left at $1$ existing), and then extended to all of $(0,2)$ as noted.
• by smooth I mean $C^\infty$. Looking at the second derivative at $x=pi/4$ (or I guess $1$ in your example) I think gives a nontrivial restriction on the function at that point, $f''-(f')^2=0$. Higher derivatives give still more restrictions, and I'm wondering if $tan(x)$ and the constant function are the unique solutions satisfying all these smoothness conditions. – asperanz Dec 1 '14 at 0:56
• @asperanz $C^\infty$ doesn't restrict things much - you can use a smooth bump function with finite support to ensure that all derivatives work at the midpoint (and in fact, even say that all derivatives are zero there) and still get arbitrary behavior away from there. – Steven Stadnicki Dec 1 '14 at 3:31
• @StevenStadnicki okay, I see. However, I have to choose the $C^\infty$ function such that its power series works at the midpoint. Setting all the derivatives to zero is like making it locally the constant function, and I could also locally make it match a $\tan$ function. Maybe I should have asked for how many analytic functions satisfy the requirement, since each analytic function would characterize the type of power series the $C^\infty$ function can have at the midpoint. – asperanz Dec 1 '14 at 4:10